A Randomized Bi-Objective Online Algorithm

In this section, a randomized algorithm for the bi-objective time series search problem is presented and analyzed by worst-component competitive analysis. As for the single-objective time series search problem, the competitive ratio is reduced by randomization.

Note that we consider the bi-objective time series search problem, i.e., n= 2. Addition-ally, we consider only the competitive ratio, rather than the strong competitive ratio.

Competitive randomized algorithms for the multi-objective time series search problem with n≥ 3 are subject to further research, as well as strongly competitive randomized algorithms for n≥2.

We follow the same approach as presented for the single-objective case in (El-Yaniv et al., 2001) and assume that ^Mi/mi = 2^ki for k_i ∈ N+, i = 1,2. Without loss of generality, suppose that k₁ ≥k₂. For the fluctuation ratio ϕof the multi-objective time series search problem we then haveϕ=^(M1M2)/(m1m2)= 2^k, wherek=k₁+k₂. Consider the deterministic algorithmsrpp-prodl for l= 0, . . . , k−1:

Algorithm 15:Multi-objective reservation price policy rpp-prodl.

1 fort= 1, . . . , T do

2 Acceptr_t= p¹_t, p²_t|

if p¹_t ·p²_t ≥m₁m₂2^l.

The randomized algorithmexpo-multchooses strategyrpp-prodl,l= 0, . . . , k−1, with probability¹/k each before the first request is revealed.

Theorem 3.3.8. For k1 ≥k2, the competitive ratio with respect tof1 of expo-multis given by

R^f1(expo-mult) = log(ϕ).

Proof. The idea for the worst case sequence presented to the online player by the adver-sary is analogous to the worst case sequence in the single-objective variant: The efficient offline solution for the adversary is supposed to be given by a request r_j+1 = p¹_j, p²_j| with p¹_t ·p²_t just below the threshold m₁m₂2^j+1 for some j. This request will not be accepted by the online player when a policy rpp-prodl with l ≥ j+ 1 is selected. In order to prevent the online player from accepting this request in case a policyrpp-prodl

withl≤j is selected, the adversary initially presents requests r_i = p¹_i, p²_i|

that match the threshold for policyrpp-prodl, i.e., withp¹_i·p²_i =m1m22ⁱ fori= 0, . . . , j. This way, the expected value of the request accepted by the online player is minimized. Choosing the index j which maximizes the competitive ratio then gives the worst case instance.

Note that, in order to maximize the competitive ratio (rather than the strong com-petitive ratio), the requests triggering the thresholds m₁m₂2ⁱ with i ≤ j have to be chosen such that the request r_j+1 remains the only efficient solution. Otherwise, the algorithm’s value would also be compared to the other efficient solutions leading to a smaller competitive ratio. See also Figures 3.5 and 3.6 for an illustration of the worst case instance.

Let r_j+1 = p¹_j+1, p²_j+1|

be an efficient offline solution and let j be an integer such that

m₁m₂2^j ≤p¹_j+1·p²_j+1< m₁m₂2^j+1.

For the maximal choice of j = k, we have pⁱ_j+1 = M_i, i = 1, . . . , n, due to bounded prices and m12^k1 ·m22^k2 =m1m22^k = M1M2. This choice is not advantageous for the adversary, since the online player obtains this price with probability ¹/k and any other choice forjgives the adversary the possibility to prevent the online player from obtaining the best price.

So, for any particular choice of j < k, the adversary selects p¹_j+1·p²_j+1 arbitrarily close tom₁m₂2^j+1, since, thereby, the adversary’s return is increased whereas the return of the online player remains unchanged. Therefore,

p¹_j+1·p²_j+1=m₁m₂2^j+1−, >0. (3.17) Ignoring the, efficient offline solutions are given by

r_j+1= p¹_j+1, p²_j+1|

=

x_j+1, ^m1m_x^22j+1

j+1

_|

for x_j+1 feasible. (3.18) The feasible values forx_j+1 are restricted byx_j+1∈[m₁, M₁]and

m₁m₂2^j+1

x_j+1 ∈[m₂, M₂] ⇔ m₁m₂2^j+1

M₂ ≤x_j+1 ≤m₁2^j+1.

p¹_t p²_t

m12⁰m12¹m12² m12³ m12⁴ m12⁵=M1

m22⁰ m₂2¹ m22² m22³=M2

p¹_t·p²_t=m₁m₂2⁶ p¹_t·p²_t=m₁m₂2⁷

rsr7

ǫ

rsr₆

rsr₅

rsr4

rsr3

rsr2

rsr1

Figure 3.5: Example for a worst case sequence in Case 1.1: k1 = 5, k2 = 3, j = 6, and x₇=m₁2⁴, the squares mark the requests.

Since M₂=m₂2^k2, we havex_j+1 ∈ Ij+1 with Ij+1:=h

maxn

m₁, m₁2^j+1−k2o ,min

M₁, m₁2^j+1 i

For the following analysis, we consider three cases with respect to the relation of j,k1, and k₂:

Case 1: j+ 1≥k₁

By assumption, we havek₁ ≥k₂, and, due toj+1≥k₁, we havej+1≥k₂. Therefore, m₁2^j+1−k2 ≥m₁ and M₁ =m₁2^k1 ≤m₁2^j+1. Thus, the intervalIj+1 is given by

Ij+1 =h

m₁2^j+1−k2, M₁i .

Due to (3.18), the competitive ratio with respect tof1(c) = maxi=1,...,nciis maximized by choosing the minimal or maximal feasible value forx_j+1, i.e.,x_j+1=m₁2^j+1−k2 or x_j+1 =M₁:

Case 1.1: x_j+1 =m₁2^j+1−k2

Consider the minimal feasible value for xj+1, i.e., xj+1 = m12^j+1−k2 (see r7 in Figure 3.5). Then, r_j+1 is given by r_j+1 = m₁2^j+1−k2, m₂2^k2_|

. Request r_j+1 will not be accepted by the online player when a policy rpp-prodl with l ≥ j+ 1 is selected. Note that for l = j+ 1, the request rj+1 is not accepted by the online player sincep¹_j+1·p²_j+1=m₁m₂2^j+1−(see (3.17), we ignored theto improve the readability).

In order to prevent the online player from accepting this request rj+1 in case a policyrpp-prodlwith0≤l≤j is selected, the adversary initially presents requests r_i = p¹_i, p²_i|

with p¹_i ·p²_i = m₁m₂2ⁱ for i = 0, . . . , j (see also Figure 3.5). Thus, requestri is given by

r_i = p¹_i, p²_i|

=

xi, ^m1m_x²²ⁱ

i

_|

, x_i ∈ Ii.

The interval Ii of feasible values for x_i is in this case determined by the following conditions: In order to obtain a feasible request, we need to guaranteex_i ∈[m₁, M₁] and

m₁m₂2ⁱ

x_i ∈[m2, M2] ⇔ m12^i−k2 ≤xi ≤m12ⁱ.

Furthermore, we have to ensure thatri 5rj+1, i.e., thatridoes not represent another efficient solution. Otherwise, the algorithm’s value would also be compared to this efficient solution leading to a smaller competitive ratio (see Definition 3.2.1). Hence, we havexi ≤m12^j+1−k2 and

m₁m₂2ⁱ

x_i ≤m₂2^k2 ⇔ x_i ≥m₁2^i−k2.

Since j+ 1 ≤ k = k₁ +k₂ and, hence, M₁ = m₁2^k1 ≥ m₁2^j+1−k2, the interval of feasible valuesIi is given by

Ii =h maxn

m₁, m₁2^i−k2o

,minn

m₁2ⁱ, m₁2^j+1−k2oi

. (3.19)

x_j+1is chosen as theminimal feasible value in order to maximize the second compo-nentp²_j+1 for the adversary. Thus, we are looking for themaximal feasible value for x_i in order to minimize the second component p²_i for the online player and, hence, maximize the competitive ratio with respect tof1.

Due to (3.19), we have xi ≤m12ⁱ for i= 0, . . . , j+ 1−k2 and xi ≤m12^j+1−k2 for i=j+ 2−k₂, . . . , j. Thus,r_i is chosen as

ri = m12ⁱ, m2_|

for i= 0, . . . , j+ 1−k2, and r_i = m₁2^j+1−k2, m₂2^i−j−1+k2_|

for i=j+ 2−k₂, . . . , j.

For the expected value of thei-th component ofalgwith respect to its randomized

decisions, denoted by E[algi], we have

The competitive ratio with respect tof₁ is then given by

k≥maxj+1≥k1

Note that the continuous maximum with respect tojis obtained forj=k−2+_log(2)¹ . However, for integer j≤k−1, the maximum is given by j=k−1.

Case 1.2: x_j+1 =M₁

Now, consider the maximal value for x_j+1, i.e., x_j+1 = M₁ (see r₇ in Figure 3.6).

Then,r_j+1 is given byr_j+1 = M₁, m₂2^j+1−k1_|

. By the same argumentation as in the previous case, the adversary previously presents requests

ri = p¹_i, p²_i|

where Ii is again the set of feasible values for xi (see also Figure 3.6). In order to obtain a feasible request, we need to guaranteex_i∈[m₁, M₁]and

p¹_t feasible valuesIi is given by

Ii =h

Sincexj+1 is in this case chosen as themaximal feasible value in order to maximize the first component p¹_j+1 for the adversary, we are looking for theminimal feasible value for x_i in order to minimize the first component p¹_i for the online player and, hence, maximize the competitive ratio with respect tof₁. Thus,r_i is given by

ri = m1, m22ⁱ_|

for i= 0, . . . , j+ 1−k1, and r_i = m₁2^i−j−1+k1, m₂2^j+1−k1_|

for i=j+ 2−k₁, . . . , j.

For the expected value of alg we then have E[alg1] = 1

= m₂ k

2^j−k1+2−1 + (k₁−1) 2^j+1−k1 +k−j−1

= m₂ k

(k₁+ 1)2^j+1−k1+k−j−2 . The competitive ratio with respect tof1 is then given by

k≥maxj+1≥k1

max

M₁

E[alg1], m₂2^j+1−k1 E[alg2]

= max

k≥j+1≥k1

max

k2^k1

2^k1+k₂−1, k2^j+1−k1

(k₁+ 1)2^j+1−k1 +k−j−2

(j+1=k)

= max

( log(ϕ) 1 +^k2−1

2^k1

, log(ϕ) k₁+ 1− ₂¹k2

)

. (3.22)

Note that the competitive ratio is symmetrical to the competitive ratio in Case 1.1 with respect to k₁ andk₂.

Case 2: j+ 1≤k₂

By assumption, we havek₁ ≥k₂, and, due toj+1≤k₂, we havej+1≤k₁. Therefore, m₁ ≥m₁2^j+1−k2 and m₁2^j+1 ≤m₁2^k1 =M₁. Thus, the intervalIj+1 is given by

Ij+1 =

m₁, m₁2^j+1 .

Due to (3.18), the competitive ratio with respect to f1 is maximized by choosing the minimal or maximal feasible value for x_j+1, i.e., x_j+1 = m₁ or x_j+1 = m₁2^j+1. We proceed with the analysis as in Case 1:

Case 2.1: xj+1 =m1

Consider the minimal feasible value for x_j+1, i.e., x_j+1 = m₁. Then, r_j+1 is given by r_j+1= m₁, m₂2^j+1_|

. By the same argumentation as in the previous cases, the adversary previously presents requests

r_i = p¹_i, p²_i|

=

xi, ^m1_x^m22i

i

_|

, x_i∈ Ii.

The intervalIiis determined byx_i∈[m₁, M₁],m₁2^i−k2 ≤x_i ≤m₁2ⁱ, andr_i5r_j+1, i.e., m₁2^i−j−1 ≤x_i ≤m₁ (for a detailed explanation see the previous cases). Conse-quently, Ii is given byIi = [m₁, m₁], i.e., x_i =m₁ and we have

r_i = m₁, m₂2ⁱ_|

for i= 0, . . . , j.

For the expected value of alg we then have The competitive ratio with respect tof₁ is then given by

j+1max≤k2

. As seen in Case 1.1 and Case 1.2, this case is symmetrical with respect tok₁ andk₂ to Case 2.1. Consequently, we have for the expected value of alg

E[alg1] = m1

k 2^j+1+k−j−2 , E[alg2] =m₂,

and the competitive ratio with respect tof₁ is given by

j+1≤kmax2

Due to (3.18), the competitive ratio with respect tof1 is maximized by choosing the minimal or maximal feasible value forx_j+1, i.e., x_j+1 =m₁2^j+1−k2 or x_j+1 =m₁2^j+1. These cases have been analyzed before, see Case 1.1 and Case 2.2.

Now, we bring together the competitive ratios of the different cases analyzed above.

By (3.20), (3.22), (3.23), and (3.24), the competitive ratio with respect to f₁ of expo-mult is given by

R^f1(expo-mult) = max

( log(ϕ) k₂+ 1−₂¹k1

, log(ϕ) 1 +^k1−1

2^k2

, log(ϕ) 1 +^k2−1

2^k1

, log(ϕ) k₁+ 1− ₂¹k2

) . Then, we have

1 +k₂−1

2^k1 ≤k₂+ 1− 1

2^k1 and 1 +k₁−1

2^k2 ≤k₁+ 1− 1 2^k2, and, due to k1 ≥k2,

k₂−1

2^k1 ≤ k₁−1 2^k2 . Therefore, we have

R^f1(expo-mult) = log(ϕ) 1 +^k2−1

2^k1

≤log(ϕ).

Im Dokument Online Resource Management (Seite 77-85)